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Alexus [3.1K]
4 years ago
11

Find two z values, one positive and one negative, that are equidistant from the mean so that the areas in the two tails add to t

he following values. a. 5% b, 10% c. 196
Mathematics
1 answer:
levacccp [35]4 years ago
3 0
I think it would be the answer is b.
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Given that sin(x)=7/11, find cos(90-x). A. 4/11 B. 7/11 C. 11/7 D. 11/4
Contact [7]
Based on the given equation above, get the value of x first so that this value can be substituted to the second equation. Value of x can be obtained by taking the arc sin of 7/11 which gives an answer of 39.52 degrees. Substituting this value to the second equation cos (90-39.52) gives an answer equals to 7/11
5 0
3 years ago
What is the length of the hypotenuse? If necessary, round to the nearest tenth.
Papessa [141]

<u>Answer:</u>

  • The value of c is 2√2 or about 2.8.

<u>Step-by-step explanation:</u>

<u>We can find the hypotenuse (c) with the help of Pythagoras Theorem.</u>

  • => c² = 2² + 2²
  • => c² = 4 + 4
  • => c² = 8
  • => c = √8
  • => c = √2 x 2 x 2
  • => c = 2√2

<u>If you want further simplification, seek for explanation below.</u>

  • => c = 2√2
  • => c = 2 x √2
  • => c = 2 x 1.414
  • => c = 2.828 = 2.8 (Estimated)

Hence, <u>the value of c is 2√2 or about 2.8.</u>

Hoped this helped.

BrainiacUser1357

6 0
2 years ago
Can some one please help me
UNO [17]

Answer: The answer is 5.

3 0
3 years ago
Which property can be used to expand the expression -2(3/4*+7)?
nika2105 [10]

Answer:

C. The distributive property

Step-by-step explanation:

The distributive property of binary activities is generalized in mathematics by the distributive law of Boolean algebra and elementary algebra. Distribution applies, in propositional logic, to two clear substitution laws. The rules allow one to reformulate, within logical facts, conjunctions and disjunctions.

7 0
3 years ago
Math:<br> Please help me I don’t understand and I need to do corrections.
GREYUIT [131]
So, you had done everything right so far (other than squaring the 2), but that was only half of the question.

to find the least common multiple, you need to first figure out what the prime factors have in common.
{2}^{2}  \times 3 \times 5 \\ and \\  {2}^{2}  \times  {3}^{2}  \times 5 \times 7
each have two twos. both have one 5, so we know our answer will look something like
{2}^{2}  \times 5 \times other \: stuff
now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.

one has one 3 and one has two, so we need two threes. now our equation is
{2}^{2}  \times {3}^{2}  \times 5 \times stuff

what's the only number we have to deal with? 7...

how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes
4 0
3 years ago
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